Question: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $n \neq 0$. $k = \dfrac{2n(3n - 2)}{-9} \div \dfrac{3n - 2}{-4} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $k = \dfrac{2n(3n - 2)}{-9} \times \dfrac{-4}{3n - 2} $ When multiplying fractions, we multiply the numerators and the denominators. $k = \dfrac{ 2n(3n - 2) \times -4 } { -9 \times (3n - 2) } $ $ k = \dfrac {-4 \times 2n(3n - 2)} {-9 (3n - 2)} $ $ k = \dfrac{-8n(3n - 2)}{-9(3n - 2)} $ We can cancel the $3n - 2$ so long as $3n - 2 \neq 0$ Therefore $n \neq \dfrac{2}{3}$ $k = \dfrac{-8n \cancel{(3n - 2})}{-9 \cancel{(3n - 2)}} = -\dfrac{8n}{-9} = \dfrac{8n}{9} $